Quadrilaterals-Activity-Mid-point theorem

Objectives

1. To guide and facilitate student exploration of the properties of triangles, specifically – a segment connecting mid-points of two sides of a triangle will be parallel to the third side and its length will be half of the third side.
2. To introduce theorems associated with triangles
3. To demonstrate the steps of logical proof and the processes
4. Understand properties of triangles

Estimated Time

One period

Prerequisites/Instructions, prior preparations, if any

An understanding of the basic elements of geometry.  Students are familiar with angles and parallel lines, and have been introduced to the concepts of parallelogram

Materials/ Resources needed

Digital - Computer, Geogebra application, projector. Geogebra files- Mid-point theorem1.ggb and Mid-point theorem2.ggb

Non digital - worksheet and pencil.

Process (How to do the activity)

Work sheet

Each group member will construct on her / his notebook, using pencil, scale, protractor, and compass a triangle, with the measures provided. You can make chits that have triangle measures.  You can even repeat them.  For a group give the following. One acute.  One obtuse. One equilateral. One isosceles. One right.  Provide only measures, for e.g.

1. Draw a triangle with following measures - BC = 6 cms, AC = 7 cms, AB = 8 cms (BC as the base)
2. Draw a triangle with following measures BC = 8 cms, angle B = 110, AB = 6 cms (BC as the base)
3. Draw a triangle with following measures BC = 10 cms, angle B = 60, angle C = 60 (BC as the base)
4. Draw a triangle with following measures BC = 12 cms, angle B = 90, AC = 13 cms (BC as the base)
5. Draw a triangle with following measures BC = 8 cms, angle B = 40, angle C = 40 (BC as the base)
6. Draw a triangle with following measures BC = 6 cms, angle B = 110, AB = 5 cms (BC as the base)
7. Draw a triangle with following measures BC = 8 cms, angle B = 60, angle C = 40 (BC as the base)

Each group gets this set and each member picks one chit and makes a sketch

1. Students should plot the mid-point of two sides (Segment AB, and AC, as D and E) and connect these with a line segment (Join these mid-points D and E.  Label it Segment DE). Show Mid-point theorem1.ggb step by step.
   1. They should measure the length of this segment DE and length of the third side BC and check if there is any relationship. They should measure the angles formed at the two vertices B and C connecting the third side, with the two angles formed on the two mid-points, D and E. For this each student will make three tables, each table  will have the following columns -
      1. Table for vertex A + Sides from the vertex,  Third side, Midpoint segment name, midpoint segment length, comparison with the side length.
      2. Angle formed at midpoint 1, angle at vertex B, angle at vertex C (students should name the angle and measure). Each student completes it for each vertex.

Vertex	Base	Length of base	Length of side 1	Length of side 2	Length of segment connecting mid-points	Relationship between 5 and 6	Angle at vertex 2	Angle at vertex 3	Angle at D	Angle at E	Relationship between 8 and 10, 9 and 11
1	2	3	4	5	6	7	8	9	10	11	12
A	BC		AC=	AB=	DE=
B	AC		BC=	AB=	DE=
C	AB		BC=	AC=	DE=

1. 1. 1. Ask them to see if there's any relationship between the  parameters. they've measured.
2. Question them if there is any relationship between the two segment lengths and the measures of the two pairs of angles.
   1. Ask them why these relationships are true across different constructions.
3. Prove the theorem, using Mid-point theorem2.ggb
   1. In △ ABC, D and E are the midpoints of sides AB and AC respectively.  D and E are joined.
   2. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1/2 BC.
   3. Construction: Extend line segment DE to F such that DE = EF. Draw segment CF.
   4. Proof: In △ ADE and △ CFE AE = EC  (given)
   5. ∠AED = ∠CEF (vertically opposite angles)
   6. DE = EF   (construction)
   7. Hence △ ADE ≅ △ CFE by SAS congruence rule.
      1. Therefore, ∠ADE = ∠CFE  (by CPCT) and ∠DAE = ∠FCE (by CPCT) AD = CF (by CPCT).
4. ∠ADE and ∠CFE are alternate interior angles,
5. (AB and CF are 2 lines intersected by transversal DF).
6. ∠DAE and ∠FCE are alternate interior angles,
7. (AB and CF are 2 lines intersected by transversal AC).
   1. Therefore, AB ∥∥ CF. So - BD ∥∥ CF.
8. BD = CF (since AD = BD and it is proved above that AD = CF).
   1. Thus, BDFC is a parallelogram.
9. By the properties of parallelogram, we have DF ∥∥ BC DF = BC DE ∥∥ BC.
   1. DE = 1/2BC  (DE = EF by construction)

Evaluation at the end of the activity

1. Would this theorem apply for right angled and obtuse-angled triangles?